Correct Answer - Option 4 :
\(\frac{7}{{12}}\)
Concept:
Total Probability Theorem:
Let events C1, C2 . . . Cn form partitions of the sample space S, where all the events have a non-zero probability of occurrence. For any event, A associated with S, according to the total probability theorem,
\(P\left( A \right) = \mathop \sum \limits_{k = 0}^n P\left( {{C_k}} \right)P\left( {A/{C_k}} \right)\)
Calculation:
Bag 1: 4 White, 2 Black
Bag 2: 3 White, 3 Black
Let P(B1) is the probability of selecting bag 1
P(B2) is the probability of selecting bag 2
P(W) is the probability of selecting white
P(W/B1) is the probability of selecting white given that the selected bag is Bag 1
P(W/B2) is the probability of selecting white given that the selected bag is Bag 2
As there are two bags, the probability selecting any of the bag is same and it is equal to
\(P\left( {B1} \right) = P\left( {B2} \right) = \frac{1}{2}\)
If the selected bag is Bag1, the probability of selecting white is
As there are 4 whites out of total 6 balls
\(P\left( {W/B1} \right) = \frac{4}{6} = \frac{2}{3}\)
If the selected bag is Bag2, the probability of selecting white is
As there are 3 whites out of total 6 balls
\(P\left( {W/B2} \right) = \frac{3}{6} = \frac{1}{2}\)
Now, from the total probability theorem
\(P\left( W \right) = P\left( {W/B1} \right)P\left( {B1} \right) + P\left( {W/B2} \right)P\left( {B2} \right)\)
\( = \frac{2}{3} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{7}{{12}}\)